Problem: $h(x) = -x$ $g(n) = n^{3}+3n^{2}+n-2+3(h(n))$ $f(t) = 6t-1+5(h(t))$ $ g(f(-2)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(-2)$ . Then we'll know what to plug into the outer function. $f(-2) = (6)(-2)-1+5(h(-2))$ To solve for the value of $f$ , we need to solve for the value of $h(-2)$ $h(-2) = -(-2)$ $h(-2) = 2$ That means $f(-2) = (6)(-2)-1+(5)(2)$ $f(-2) = -3$ Now we know that $f(-2) = -3$ . Let's solve for $g(f(-2))$ , which is $g(-3)$ $g(-3) = (-3)^{3}+3(-3)^{2}-3-2+3(h(-3))$ To solve for the value of $g$ , we need to solve for the value of $h(-3)$ $h(-3) = -(-3)$ $h(-3) = 3$ That means $g(-3) = (-3)^{3}+3(-3)^{2}-3-2+(3)(3)$ $g(-3) = 4$